I have a tall table which contains up to 10 values per group. How can I transform this table into a wide format i.e. add 2 columns where these resemble the value smaller or equal to a threshold?
I want to find the maximum per group, but it needs to be smaller than a specified value like:
min(max('value1), lit(5)).over(Window.partitionBy('grouping))
However min()will only work for a column and not for the Scala value which is returned from the inner function?
The problem can be described as:
Seq(Seq(1,2,3,4).max,5).min
Where Seq(1,2,3,4) is returned by the window.
How can I formulate this in spark sql?
edit
E.g.
+--------+-----+---------+
|grouping|value|something|
+--------+-----+---------+
| 1| 1| first|
| 1| 2| second|
| 1| 3| third|
| 1| 4| fourth|
| 1| 7| 7|
| 1| 10| 10|
| 21| 1| first|
| 21| 2| second|
| 21| 3| third|
+--------+-----+---------+
created by
case class MyThing(grouping: Int, value:Int, something:String)
val df = Seq(MyThing(1,1, "first"), MyThing(1,2, "second"), MyThing(1,3, "third"),MyThing(1,4, "fourth"),MyThing(1,7, "7"), MyThing(1,10, "10"),
MyThing(21,1, "first"), MyThing(21,2, "second"), MyThing(21,3, "third")).toDS
Where
df
.withColumn("somethingAtLeast5AndMaximum5", max('value).over(Window.partitionBy('grouping)))
.withColumn("somethingAtLeast6OupToThereshold2", max('value).over(Window.partitionBy('grouping)))
.show
returns
+--------+-----+---------+----------------------------+-------------------------+
|grouping|value|something|somethingAtLeast5AndMaximum5| somethingAtLeast6OupToThereshold2 |
+--------+-----+---------+----------------------------+-------------------------+
| 1| 1| first| 10| 10|
| 1| 2| second| 10| 10|
| 1| 3| third| 10| 10|
| 1| 4| fourth| 10| 10|
| 1| 7| 7| 10| 10|
| 1| 10| 10| 10| 10|
| 21| 1| first| 3| 3|
| 21| 2| second| 3| 3|
| 21| 3| third| 3| 3|
+--------+-----+---------+----------------------------+-------------------------+
Instead, I rather would want to formulate:
lit(Seq(max('value).asInstanceOf[java.lang.Integer], new java.lang.Integer(2)).min).over(Window.partitionBy('grouping))
But that does not work as max('value) is not a scalar value.
Expected output should look like
+--------+-----+---------+----------------------------+-------------------------+
|grouping|value|something|somethingAtLeast5AndMaximum5|somethingAtLeast6OupToThereshold2|
+--------+-----+---------+----------------------------+-------------------------+
| 1| 4| fourth| 4| 7|
| 21| 1| first| 3| NULL|
+--------+-----+---------+----------------------------+-------------------------+
edit2
When trying a pivot
df.groupBy("grouping").pivot("value").agg(first('something)).show
+--------+-----+------+-----+------+----+----+
|grouping| 1| 2| 3| 4| 7| 10|
+--------+-----+------+-----+------+----+----+
| 1|first|second|third|fourth| 7| 10|
| 21|first|second|third| null|null|null|
+--------+-----+------+-----+------+----+----+
The second part of the problem remains that some columns might not exist or be null.
When aggregating to arrays:
df.groupBy("grouping").agg(collect_list('value).alias("value"), collect_list('something).alias("something"))
+--------+-------------------+--------------------+
|grouping| value| something|
+--------+-------------------+--------------------+
| 1|[1, 2, 3, 4, 7, 10]|[first, second, t...|
| 21| [1, 2, 3]|[first, second, t...|
+--------+-------------------+--------------------+
The values are already next to each other, but the right values need to be selected. This is probably still more efficient than a join or window function.
Would be easier to do in two separate steps - calculate max over Window, and then use when...otherwise on result to produce min(x, 5):
df.withColumn("tmp", max('value1).over(Window.partitionBy('grouping)))
.withColumn("result", when('tmp > lit(5), 5).otherwise('tmp))
EDIT: some example data to clarify this:
val df = Seq((1, 1),(1, 2),(1, 3),(1, 4),(2, 7),(2, 8))
.toDF("grouping", "value1")
df.withColumn("result", max('value1).over(Window.partitionBy('grouping)))
.withColumn("result", when('result > lit(5), 5).otherwise('result))
.show()
// +--------+------+------+
// |grouping|value1|result|
// +--------+------+------+
// | 1| 1| 4| // 4, because Seq(Seq(1,2,3,4).max,5).min = 4
// | 1| 2| 4|
// | 1| 3| 4|
// | 1| 4| 4|
// | 2| 7| 5| // 5, because Seq(Seq(7,8).max,5).min = 5
// | 2| 8| 5|
// +--------+------+------+
Related
Logic and columnIn Pyspark DataFrame consider a column like [1,2,3,4,1,2,1,1,2,3,1,2,1,1,2]. Pyspark Column
create a new column to increment value when value resets to 1.
Expected output is[1,1,1,1,2,2,3,4,4,4,5,5,6,7,7]
i am bit new to pyspark, if anyone can help me it would be great for me.
written the logic as like below
def sequence(row_num):
results = [1, ]
flag = 1
for col in range(0, len(row_num)-1):
if row_num[col][0]>=row_num[col+1][0]:
flag+=1
results.append(flag)
return results
but not able to pass a column through udf. please help me in this
Your Dataframe:
df = spark.createDataFrame(
[
('1','a'),
('2','b'),
('3','c'),
('4','d'),
('1','e'),
('2','f'),
('1','g'),
('1','h'),
('2','i'),
('3','j'),
('1','k'),
('2','l'),
('1','m'),
('1','n'),
('2','o')
], ['group','label']
)
+-----+-----+
|group|label|
+-----+-----+
| 1| a|
| 2| b|
| 3| c|
| 4| d|
| 1| e|
| 2| f|
| 1| g|
| 1| h|
| 2| i|
| 3| j|
| 1| k|
| 2| l|
| 1| m|
| 1| n|
| 2| o|
+-----+-----+
You can create a flag and use a Window Function to calculate the cumulative sum. No need to use an UDF:
from pyspark.sql import Window as W
from pyspark.sql import functions as F
w = W.partitionBy().orderBy('label').rowsBetween(Window.unboundedPreceding, 0)
df\
.withColumn('Flag', F.when(F.col('group') == 1, 1).otherwise(0))\
.withColumn('Output', F.sum('Flag').over(w))\
.show()
+-----+-----+----+------+
|group|label|Flag|Output|
+-----+-----+----+------+
| 1| a| 1| 1|
| 2| b| 0| 1|
| 3| c| 0| 1|
| 4| d| 0| 1|
| 1| e| 1| 2|
| 2| f| 0| 2|
| 1| g| 1| 3|
| 1| h| 1| 4|
| 2| i| 0| 4|
| 3| j| 0| 4|
| 1| k| 1| 5|
| 2| l| 0| 5|
| 1| m| 1| 6|
| 1| n| 1| 7|
| 2| o| 0| 7|
+-----+-----+----+------+
I will expose my problem based on the initial dataframe and the one I want to achieve:
val df_997 = Seq [(Int, Int, Int, Int)]((1,1,7,10),(1,10,4,300),(1,3,14,50),(1,20,24,70),(1,30,12,90),(2,10,4,900),(2,25,30,40),(2,15,21,60),(2,5,10,80)).toDF("policyId","FECMVTO","aux","IND_DEF").orderBy(asc("policyId"), asc("FECMVTO"))
df_997.show
+--------+-------+---+-------+
|policyId|FECMVTO|aux|IND_DEF|
+--------+-------+---+-------+
| 1| 1| 7| 10|
| 1| 3| 14| 50|
| 1| 10| 4| 300|
| 1| 20| 24| 70|
| 1| 30| 12| 90|
| 2| 5| 10| 80|
| 2| 10| 4| 900|
| 2| 15| 21| 60|
| 2| 25| 30| 40|
+--------+-------+---+-------+
Imagine I have partitioned this DF by the column policyId and created the column row_num based on it to better see the Windows:
val win = Window.partitionBy("policyId").orderBy("FECMVTO")
val df_998 = df_997.withColumn("row_num",row_number().over(win))
df_998.show
+--------+-------+---+-------+-------+
|policyId|FECMVTO|aux|IND_DEF|row_num|
+--------+-------+---+-------+-------+
| 1| 1| 7| 10| 1|
| 1| 3| 14| 50| 2|
| 1| 10| 4| 300| 3|
| 1| 20| 24| 70| 4|
| 1| 30| 12| 90| 5|
| 2| 5| 10| 80| 1|
| 2| 10| 4| 900| 2|
| 2| 15| 21| 60| 3|
| 2| 25| 30| 40| 4|
+--------+-------+---+-------+-------+
Now, for each window, if the value of aux is 4, I want to set the value of IND_DEF column for that register to the column FEC_MVTO for this register on until the end of the window.
The resulting DF would be:
+--------+-------+---+-------+-------+
|policyId|FECMVTO|aux|IND_DEF|row_num|
+--------+-------+---+-------+-------+
| 1| 1| 7| 10| 1|
| 1| 3| 14| 50| 2|
| 1| 300| 4| 300| 3|
| 1| 300| 24| 70| 4|
| 1| 300| 12| 90| 5|
| 2| 5| 10| 80| 1|
| 2| 900| 4| 900| 2|
| 2| 900| 21| 60| 3|
| 2| 900| 30| 40| 4|
+--------+-------+---+-------+-------+
Thanks for your suggestions as I am very stuck in here...
Here's one approach: First left-join the DataFrame with its aux == 4 filtered version, followed by applying Window function first to backfill nulls with the wanted IND_DEF values per partition, and finally conditionally recreate column FECMVTO:
import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions._
import spark.implicits._
val df = Seq(
(1,1,7,10), (1,10,4,300), (1,3,14,50), (1,20,24,70), (1,30,12,90),
(2,10,4,900), (2,25,30,40), (2,15,21,60), (2,5,10,80)
).toDF("policyId","FECMVTO","aux","IND_DEF")
val win = Window.partitionBy("policyId").orderBy("FECMVTO").
rowsBetween(Window.unboundedPreceding, 0)
val df2 = df.
select($"policyId", $"aux", $"IND_DEF".as("IND_DEF2")).
where($"aux" === 4)
df.join(df2, Seq("policyId", "aux"), "left_outer").
withColumn("IND_DEF3", first($"IND_DEF2", ignoreNulls=true).over(win)).
withColumn("FECMVTO", coalesce($"IND_DEF3", $"FECMVTO")).
show
// +--------+---+-------+-------+--------+--------+
// |policyId|aux|FECMVTO|IND_DEF|IND_DEF2|IND_DEF3|
// +--------+---+-------+-------+--------+--------+
// | 1| 7| 1| 10| null| null|
// | 1| 14| 3| 50| null| null|
// | 1| 4| 300| 300| 300| 300|
// | 1| 24| 300| 70| null| 300|
// | 1| 12| 300| 90| null| 300|
// | 2| 10| 5| 80| null| null|
// | 2| 4| 900| 900| 900| 900|
// | 2| 21| 900| 60| null| 900|
// | 2| 30| 900| 40| null| 900|
// +--------+---+-------+-------+--------+--------+
Columns IND_DEF2, IND_DEF3 are kept only for illustration (and can certainly be dropped).
#I believe below can be solution for your issue
Considering input_df is your input dataframe
//Step#1 - Filter rows with IND_DEF = 4 from input_df
val only_FECMVTO_4_df1 = input_df.filter($"IND_DEF" === 4)
//Step#2 - Filling FECMVTO value from IND_DEF for the above result
val only_FECMVTO_4_df2 = only_FECMVTO_4_df1.withColumn("FECMVTO_NEW",$"IND_DEF").drop($"FECMVTO").withColumnRenamed("FECMVTO",$"FECMVTO_NEW")
//Step#3 - removing all the records from step#1 from input_df
val input_df_without_FECMVTO_4 = input_df.except(only_FECMVTO_4_df1)
//combining Step#2 output with output of Step#3
val final_df = input_df_without_FECMVTO_4.union(only_FECMVTO_4_df2)
I would like to aggregate this DataFrame and count the number of observations with a value less than or equal to the "BUCKET" field for each level. For example:
val myDF = Seq(
("foo", 0),
("foo", 0),
("bar", 0),
("foo", 1),
("foo", 1),
("bar", 1),
("foo", 2),
("bar", 2),
("foo", 3),
("bar", 3)).toDF("COL1", "BUCKET")
myDF.show
+----+------+
|COL1|BUCKET|
+----+------+
| foo| 0|
| foo| 0|
| bar| 0|
| foo| 1|
| foo| 1|
| bar| 1|
| foo| 2|
| bar| 2|
| foo| 3|
| bar| 3|
+----+------+
I can count the number of observations matching each bucket value using this code:
myDF.groupBy("COL1").pivot("BUCKET").count.show
+----+---+---+---+---+
|COL1| 0| 1| 2| 3|
+----+---+---+---+---+
| bar| 1| 1| 1| 1|
| foo| 2| 2| 1| 1|
+----+---+---+---+---+
But I want to count the number of rows with a value in the "BUCKET" field which is less than or equal to the final header after pivoting, like this:
+----+---+---+---+---+
|COL1| 0| 1| 2| 3|
+----+---+---+---+---+
| bar| 1| 2| 3| 4|
| foo| 2| 4| 5| 6|
+----+---+---+---+---+
You can achieve this using a window function, as follows:
import org.apache.spark.sql.expressions.Window.partitionBy
import org.apache.spark.sql.functions.first
myDF.
select(
$"COL1",
$"BUCKET",
count($"BUCKET").over(partitionBy($"COL1").orderBy($"BUCKET")).as("ROLLING_COUNT")).
groupBy($"COL1").pivot("BUCKET").agg(first("ROLLING_COUNT")).
show()
+----+---+---+---+---+
|COL1| 0| 1| 2| 3|
+----+---+---+---+---+
| bar| 1| 2| 3| 4|
| foo| 2| 4| 5| 6|
+----+---+---+---+---+
What you are specifying here is that you want to perform a count of your observations, partitioned in windows as determined by a key (COL1 in this case). By specifying an ordering, you are also making the count rolling over the window, thus obtaining the results you want then to be pivoted in your end results.
This is the result of applying the window function:
myDF.
select(
$"COL1",
$"BUCKET",
count($"BUCKET").over(partitionBy($"COL1").orderBy($"BUCKET")).as("ROLLING_COUNT")).
show()
+----+------+-------------+
|COL1|BUCKET|ROLLING_COUNT|
+----+------+-------------+
| bar| 0| 1|
| bar| 1| 2|
| bar| 2| 3|
| bar| 3| 4|
| foo| 0| 2|
| foo| 0| 2|
| foo| 1| 4|
| foo| 1| 4|
| foo| 2| 5|
| foo| 3| 6|
+----+------+-------------+
Finally, by grouping by COL1, pivoting over BUCKET and only getting the first result of the rolling count (anyone would be good as all of them are applied to the whole window), you finally obtain the result you were looking for.
In a way, window functions are very similar to aggregations over groupings, but are more flexible and powerful. This just scratches the surface of window functions and you can dig a little bit deeper by having a look at this introductory reading.
Here's one approach to get the rolling counts by traversing the pivoted BUCKET value columns using foldLeft to aggregate the counts. Note that a tuple of (DataFrame, Int) is used for foldLeft to transform the DataFrame as well as store the count in the previous iteration:
val pivotedDF = myDF.groupBy($"COL1").pivot("BUCKET").count
val buckets = pivotedDF.columns.filter(_ != "COL1")
buckets.drop(1).foldLeft((pivotedDF, buckets.head))( (acc, c) =>
( acc._1.withColumn(c, col(acc._2) + col(c)), c )
)._1.show
// +----+---+---+---+---+
// |COL1| 0| 1| 2| 3|
// +----+---+---+---+---+
// | bar| 1| 2| 3| 4|
// | foo| 2| 4| 5| 6|
// +----+---+---+---+---+
I'm looking for a way to rank columns of a dataframe preserving ties. Specifically for this example, I have a pyspark dataframe as follows where I want to generate ranks for colA & colB (though I want to support being able to rank N number of columns)
+--------+----------+-----+----+
| Entity| id| colA|colB|
+-------------------+-----+----+
| a|8589934652| 21| 50|
| b| 112| 9| 23|
| c|8589934629| 9| 23|
| d|8589934702| 8| 21|
| e| 20| 2| 21|
| f|8589934657| 2| 5|
| g|8589934601| 1| 5|
| h|8589934653| 1| 4|
| i|8589934620| 0| 4|
| j|8589934643| 0| 3|
| k|8589934618| 0| 3|
| l|8589934602| 0| 2|
| m|8589934664| 0| 2|
| n| 25| 0| 1|
| o| 67| 0| 1|
| p|8589934642| 0| 1|
| q|8589934709| 0| 1|
| r|8589934660| 0| 1|
| s| 30| 0| 1|
| t| 55| 0| 1|
+--------+----------+-----+----+
What I'd like is a way to rank this dataframe where tied values receive the same rank such as:
+--------+----------+-----+----+---------+---------+
| Entity| id| colA|colB|colA_rank|colB_rank|
+-------------------+-----+----+---------+---------+
| a|8589934652| 21| 50| 1| 1|
| b| 112| 9| 23| 2| 2|
| c|8589934629| 9| 21| 2| 3|
| d|8589934702| 8| 21| 3| 3|
| e| 20| 2| 21| 4| 3|
| f|8589934657| 2| 5| 4| 4|
| g|8589934601| 1| 5| 5| 4|
| h|8589934653| 1| 4| 5| 5|
| i|8589934620| 0| 4| 6| 5|
| j|8589934643| 0| 3| 6| 6|
| k|8589934618| 0| 3| 6| 6|
| l|8589934602| 0| 2| 6| 7|
| m|8589934664| 0| 2| 6| 7|
| n| 25| 0| 1| 6| 8|
| o| 67| 0| 1| 6| 8|
| p|8589934642| 0| 1| 6| 8|
| q|8589934709| 0| 1| 6| 8|
| r|8589934660| 0| 1| 6| 8|
| s| 30| 0| 1| 6| 8|
| t| 55| 0| 1| 6| 8|
+--------+----------+-----+----+---------+---------+
My current implementation with the first dataframe looks like:
def getRanks(mydf, cols=None, ascending=False):
from pyspark import Row
# This takes a dataframe and a list of columns to rank
# If no list is provided, it ranks *all* columns
# returns a new dataframe
def addRank(ranked_rdd, col, ascending):
# This assumes an RDD of the form (Row(...), list[...])
# it orders the rdd by col, finds the order, then adds that to the
# list
myrdd = ranked_rdd.sortBy(lambda (row, ranks): row[col],
ascending=ascending).zipWithIndex()
return myrdd.map(lambda ((row, ranks), index): (row, ranks +
[index+1]))
myrdd = mydf.rdd
fields = myrdd.first().__fields__
ranked_rdd = myrdd.map(lambda x: (x, []))
if (cols is None):
cols = fields
for col in cols:
ranked_rdd = addRank(ranked_rdd, col, ascending)
rank_names = [x + "_rank" for x in cols]
# Hack to make sure columns come back in the right order
ranked_rdd = ranked_rdd.map(lambda (row, ranks): Row(*row.__fields__ +
rank_names)(*row + tuple(ranks)))
return ranked_rdd.toDF()
which produces:
+--------+----------+-----+----+---------+---------+
| Entity| id| colA|colB|colA_rank|colB_rank|
+-------------------+-----+----+---------+---------+
| a|8589934652| 21| 50| 1| 1|
| b| 112| 9| 23| 2| 2|
| c|8589934629| 9| 23| 3| 3|
| d|8589934702| 8| 21| 4| 4|
| e| 20| 2| 21| 5| 5|
| f|8589934657| 2| 5| 6| 6|
| g|8589934601| 1| 5| 7| 7|
| h|8589934653| 1| 4| 8| 8|
| i|8589934620| 0| 4| 9| 9|
| j|8589934643| 0| 3| 10| 10|
| k|8589934618| 0| 3| 11| 11|
| l|8589934602| 0| 2| 12| 12|
| m|8589934664| 0| 2| 13| 13|
| n| 25| 0| 1| 14| 14|
| o| 67| 0| 1| 15| 15|
| p|8589934642| 0| 1| 16| 16|
| q|8589934709| 0| 1| 17| 17|
| r|8589934660| 0| 1| 18| 18|
| s| 30| 0| 1| 19| 19|
| t| 55| 0| 1| 20| 20|
+--------+----------+-----+----+---------+---------+
As you can see, the function getRanks() takes a dataframe, specifies the columns to be ranked, sorts them, and uses zipWithIndex() to generate an ordering or rank. However, I can't figure out a way to preserve ties.
This stackoverflow post is the closest solution I've found:
rank-users-by-column But it appears to only handle 1 column (I think).
Thanks so much for the help in advance!
EDIT: column 'id' is generated from calling monotonically_increasing_id() and in my implementation is cast to a string.
You're looking for dense_rank
First let's create our dataframe:
df = spark.createDataFrame(sc.parallelize([["a",8589934652,21,50],["b",112,9,23],["c",8589934629,9,23],
["d",8589934702,8,21],["e",20,2,21],["f",8589934657,2,5],
["g",8589934601,1,5],["h",8589934653,1,4],["i",8589934620,0,4],
["j",8589934643,0,3],["k",8589934618,0,3],["l",8589934602,0,2],
["m",8589934664,0,2],["n",25,0,1],["o",67,0,1],["p",8589934642,0,1],
["q",8589934709,0,1],["r",8589934660,0,1],["s",30,0,1],["t",55,0,1]]
), ["Entity","id","colA","colB"])
We'll define two windowSpec:
from pyspark.sql import Window
import pyspark.sql.functions as psf
wA = Window.orderBy(psf.desc("colA"))
wB = Window.orderBy(psf.desc("colB"))
df = df.withColumn(
"colA_rank",
psf.dense_rank().over(wA)
).withColumn(
"colB_rank",
psf.dense_rank().over(wB)
)
+------+----------+----+----+---------+---------+
|Entity| id|colA|colB|colA_rank|colB_rank|
+------+----------+----+----+---------+---------+
| a|8589934652| 21| 50| 1| 1|
| b| 112| 9| 23| 2| 2|
| c|8589934629| 9| 23| 2| 2|
| d|8589934702| 8| 21| 3| 3|
| e| 20| 2| 21| 4| 3|
| f|8589934657| 2| 5| 4| 4|
| g|8589934601| 1| 5| 5| 4|
| h|8589934653| 1| 4| 5| 5|
| i|8589934620| 0| 4| 6| 5|
| j|8589934643| 0| 3| 6| 6|
| k|8589934618| 0| 3| 6| 6|
| l|8589934602| 0| 2| 6| 7|
| m|8589934664| 0| 2| 6| 7|
| n| 25| 0| 1| 6| 8|
| o| 67| 0| 1| 6| 8|
| p|8589934642| 0| 1| 6| 8|
| q|8589934709| 0| 1| 6| 8|
| r|8589934660| 0| 1| 6| 8|
| s| 30| 0| 1| 6| 8|
| t| 55| 0| 1| 6| 8|
+------+----------+----+----+---------+---------+
I'll also pose an alternative:
for cols in data.columns[2:]:
lookup = (data.select(cols)
.distinct()
.orderBy(cols, ascending=False)
.rdd
.zipWithIndex()
.map(lambda x: x[0] + (x[1], ))
.toDF([cols, cols+"_rank_lookup"]))
name = cols + "_ranks"
data = data.join(lookup, [cols]).withColumn(name,col(cols+"_rank_lookup")
+ 1).drop(cols + "_rank_lookup")
Not as elegant as dense_rank() and I'm uncertain as to performance implications.
I have a dataframe(spark):
id value
3 0
3 1
3 0
4 1
4 0
4 0
I want to create a new dataframe:
3 0
3 1
4 1
Need to remove all the rows after 1(value) for each id.I tried with window functions in spark dateframe(Scala). But couldn't able to find a solution.Seems to be I am going in a wrong direction.
I am looking for a solution in Scala.Thanks
Output using monotonically_increasing_id
scala> val data = Seq((3,0),(3,1),(3,0),(4,1),(4,0),(4,0)).toDF("id", "value")
data: org.apache.spark.sql.DataFrame = [id: int, value: int]
scala> val minIdx = dataWithIndex.filter($"value" === 1).groupBy($"id").agg(min($"idx")).toDF("r_id", "min_idx")
minIdx: org.apache.spark.sql.DataFrame = [r_id: int, min_idx: bigint]
scala> dataWithIndex.join(minIdx,($"r_id" === $"id") && ($"idx" <= $"min_idx")).select($"id", $"value").show
+---+-----+
| id|value|
+---+-----+
| 3| 0|
| 3| 1|
| 4| 1|
+---+-----+
The solution wont work if we did a sorted transformation in the original dataframe. That time the monotonically_increasing_id() is generated based on original DF rather that sorted DF.I have missed that requirement before.
All suggestions are welcome.
One way is to use monotonically_increasing_id() and a self-join:
val data = Seq((3,0),(3,1),(3,0),(4,1),(4,0),(4,0)).toDF("id", "value")
data.show
+---+-----+
| id|value|
+---+-----+
| 3| 0|
| 3| 1|
| 3| 0|
| 4| 1|
| 4| 0|
| 4| 0|
+---+-----+
Now we generate a column named idx with an increasing Long:
val dataWithIndex = data.withColumn("idx", monotonically_increasing_id())
// dataWithIndex.cache()
Now we get the min(idx) for each id where value = 1:
val minIdx = dataWithIndex
.filter($"value" === 1)
.groupBy($"id")
.agg(min($"idx"))
.toDF("r_id", "min_idx")
Now we join the min(idx) back to the original DataFrame:
dataWithIndex.join(
minIdx,
($"r_id" === $"id") && ($"idx" <= $"min_idx")
).select($"id", $"value").show
+---+-----+
| id|value|
+---+-----+
| 3| 0|
| 3| 1|
| 4| 1|
+---+-----+
Note: monotonically_increasing_id() generates its value based on the partition of the row. This value may change each time dataWithIndex is re-evaluated. In my code above, because of lazy evaluation, it's only when I call the final show that monotonically_increasing_id() is evaluated.
If you want to force the value to stay the same, for example so you can use show to evaluate the above step-by-step, uncomment this line above:
// dataWithIndex.cache()
Hi I found the solution using Window and self join.
val data = Seq((3,0,2),(3,1,3),(3,0,1),(4,1,6),(4,0,5),(4,0,4),(1,0,7),(1,1,8),(1,0,9),(2,1,10),(2,0,11),(2,0,12)).toDF("id", "value","sorted")
data.show
scala> data.show
+---+-----+------+
| id|value|sorted|
+---+-----+------+
| 3| 0| 2|
| 3| 1| 3|
| 3| 0| 1|
| 4| 1| 6|
| 4| 0| 5|
| 4| 0| 4|
| 1| 0| 7|
| 1| 1| 8|
| 1| 0| 9|
| 2| 1| 10|
| 2| 0| 11|
| 2| 0| 12|
+---+-----+------+
val sort_df=data.sort($"sorted")
scala> sort_df.show
+---+-----+------+
| id|value|sorted|
+---+-----+------+
| 3| 0| 1|
| 3| 0| 2|
| 3| 1| 3|
| 4| 0| 4|
| 4| 0| 5|
| 4| 1| 6|
| 1| 0| 7|
| 1| 1| 8|
| 1| 0| 9|
| 2| 1| 10|
| 2| 0| 11|
| 2| 0| 12|
+---+-----+------+
var window=Window.partitionBy("id").orderBy("$sorted")
val sort_idx=sort_df.select($"*",rowNumber.over(window).as("count_index"))
val minIdx=sort_idx.filter($"value"===1).groupBy("id").agg(min("count_index")).toDF("idx","min_idx")
val result_id=sort_idx.join(minIdx,($"id"===$"idx") &&($"count_index" <= $"min_idx"))
result_id.show
+---+-----+------+-----------+---+-------+
| id|value|sorted|count_index|idx|min_idx|
+---+-----+------+-----------+---+-------+
| 1| 0| 7| 1| 1| 2|
| 1| 1| 8| 2| 1| 2|
| 2| 1| 10| 1| 2| 1|
| 3| 0| 1| 1| 3| 3|
| 3| 0| 2| 2| 3| 3|
| 3| 1| 3| 3| 3| 3|
| 4| 0| 4| 1| 4| 3|
| 4| 0| 5| 2| 4| 3|
| 4| 1| 6| 3| 4| 3|
+---+-----+------+-----------+---+-------+
Still looking for a more optimized solutions.Thanks
You can simply use groupBy like this
val df2 = df1.groupBy("id","value").count().select("id","value")
Here your df1 is
id value
3 0
3 1
3 0
4 1
4 0
4 0
And resultant dataframe is df2 which is your expected output like this
id value
3 0
3 1
4 1
4 0
use isin method and filter as below:
val data = Seq((3,0,2),(3,1,3),(3,0,1),(4,1,6),(4,0,5),(4,0,4),(1,0,7),(1,1,8),(1,0,9),(2,1,10),(2,0,11),(2,0,12)).toDF("id", "value","sorted")
val idFilter = List(1, 2)
data.filter($"id".isin(idFilter:_*)).show
+---+-----+------+
| id|value|sorted|
+---+-----+------+
| 1| 0| 7|
| 1| 1| 8|
| 1| 0| 9|
| 2| 1| 10|
| 2| 0| 11|
| 2| 0| 12|
+---+-----+------+
Ex: filter based on val
val valFilter = List(0)
data.filter($"value".isin(valFilter:_*)).show
+---+-----+------+
| id|value|sorted|
+---+-----+------+
| 3| 0| 2|
| 3| 0| 1|
| 4| 0| 5|
| 4| 0| 4|
| 1| 0| 7|
| 1| 0| 9|
| 2| 0| 11|
| 2| 0| 12|
+---+-----+------+